Simplify the following expression: $y = \dfrac{8x^2+13x- 6}{8x - 3}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(8)}{(-6)} &=& -48 \\ {a} + {b} &=& &=& {13} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-48$ and add them together. Remember, since $-48$ is negative, one of the factors must be negative. The factors that add up to ${13}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${16}$ $ \begin{eqnarray} {ab} &=& ({-3})({16}) &=& -48 \\ {a} + {b} &=& {-3} + {16} &=& 13 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({8}x^2 {-3}x) + ({16}x {-6}) $ Factor out the common factors: $ x(8x - 3) + 2(8x - 3)$ Now factor out $(8x - 3)$ $ (8x - 3)(x + 2)$ The original expression can therefore be written: $ \dfrac{(8x - 3)(x + 2)}{8x - 3}$ We are dividing by $8x - 3$ , so $8x - 3 \neq 0$ Therefore, $x \neq \frac{3}{8}$ This leaves us with $x + 2; x \neq \frac{3}{8}$.